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100 POINTS PRE-CALCULUS

An airplane is flying in the direction of 30 degrees South of West at 600 mph. A wind is blowing in the direction of 60 degrees North of East at 40 mph.


are the component forms for the air and wind vectors.


The ground speed, G, of the plane is given by

.

100 POINTS PRECALCULUSAn Airplane Is Flying In The Direction Of 30 Degrees South Of West At 600 Mph A Wind Is Blowing In The Direction Of 60 Degrees North Of Ea class=
100 POINTS PRECALCULUSAn Airplane Is Flying In The Direction Of 30 Degrees South Of West At 600 Mph A Wind Is Blowing In The Direction Of 60 Degrees North Of Ea class=
100 POINTS PRECALCULUSAn Airplane Is Flying In The Direction Of 30 Degrees South Of West At 600 Mph A Wind Is Blowing In The Direction Of 60 Degrees North Of Ea class=

Sagot :

Answer:

not sure for the 1st one but pretty sure 2nd one is B

Step-by-step explanation:

The ground speed (G) is given as

[tex]G = \sqrt{(600\dfrac{\sqrt3}{2} + 20)^2 + (300 + 40 \dfrac{\sqrt3}{2})^2}[/tex].

The correct potion is B.

What is a vector?

The quantity which has magnitude, direction and follows the law of vector addition is called a vector.

An airplane is flying in the direction of 30 degrees South of West at 600 mph. A wind is blowing in the direction of 60 degrees North of East at 40 mph.

It means the airplane is flying in the opposite direction of the wind.

The component of an airplane will be

[tex]\rm x = - 600\ sin 30\\\\x = - \dfrac{600}{2}= -300\\\\y = -600 \ cos30\\ \\y= -\dfrac{600 \sqrt{3}}{2}[/tex]

The component of wind will be

[tex]\rm x = 40 cos 60\\\\x = \dfrac{40}{2} = 20 \\\\y = 40sin 45\\ \\y= \dfrac{40\sqrt3}{2}[/tex]

Then the ground speed will be

[tex]G = \sqrt{(600\dfrac{\sqrt3}{2} + 20)^2 + (300 + 40 \dfrac{\sqrt3}{2})^2}[/tex]

Then the correct potion is B.

More about the vector link is given below.

https://brainly.com/question/13188123

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